Undergrad Why is the Matter+Lambda Universe Equation Equal to 1-Ω?

[Das apashanka]

for the matter+lambda universe why is Ω[Λ,o]=1-Ω[m,0]?

 

[PeterDonis]

for the matter+lambda universe why is Ω[Λ,o]=1-Ω[m,0]?

Where are you getting this equation from? Do you have a reference?

 

[kimbyd]

for the matter+lambda universe why is Ω[Λ,o]=1-Ω[m,0]?

Because the sum of the $\Omega$ terms is defined to be equal to one, and a matter+lambda universe has only matter and a cosmological constant term.

 

[Das apashanka]

will you please explain why is the net Ω is equal to 1,and net Ω consists of which components

 

[Das apashanka]

Because the sum of the $\Omega$ terms is defined to be equal to one, and a matter+lambda universe has only matter and a cosmological constant term.

will you please explain the reason for being total omega to be 1

 

[PeterDonis]

@Das apashanka , you marked this thread as "A", indicating a graduate level knowledge of the subject matter. The questions you are asking indicate that you don't have that background; accordingly, I have changed the thread level to "I".

 

[kimbyd]

will you please explain the reason for being total omega to be 1

There's no deep meaning. The parameter is just defined that way.

 

[George Jones]

for the matter+lambda universe why is Ω[Λ,o]=1-Ω[m,0]?

It isn't, necessarily.

Because the sum of the $\Omega$ terms is defined to be equal to one, and a matter+lambda universe has only matter and a cosmological constant term.

This only is true for spatially flat matter/Lambda universes, i.e., it is not true for general matter/Lambda universes.

 

[kimbyd]

This only is true for spatially flat matter/Lambda universes, i.e., it is not true for general matter/Lambda universes.

Whichever $\Omega$ values you include in your model, their sum is always identically equal to one. That's how they're defined: as density fractions. If you have matter, cosmological constant, and curvature, then it's:

$$\Omega_m + \Omega_\Lambda + \Omega_k = 1$$

 

[George Jones]

Whichever $\Omega$ values you include in your model, their sum is always identically equal to one. That's how they're defined: as density fractions. If you have matter, cosmological constant, and curvature, then it's:

$$\Omega_m + \Omega_\Lambda + \Omega_k = 1$$

Of course. As you say, this is true **by definition**. But this is not what the original poster asked about. A question equivalent to the OP's question is "Why is $\Omega_k = 0$?"

One possible answer is that the OP is looking at the (very close to) spatially flat observed universe at a time when $\Omega_r$ is negligible. One nice thing about this universe is that there is an explicit expression for the scale factor.

If, however, this an exercise in theoretical cosmology, there is no reason for

$$\Omega_m + \Omega_\Lambda = 1$$

in a matter/Lambda universe, which is what the OP asked about. So, what Peter asked is necessary,

Where are you getting this equation from? Do you have a reference?

 

[kimbyd]

Of course. As you say, this is true **by definition**. But this is not what the original poster asked about. A question equivalent to the OP's question is "Why is $\Omega_k = 0$?"

That's more or less the point I was getting at. There's no deep reason here. It's just the way the terms are defined.

But yes, there might be a more important question that is being missed here, so we might be able to offer better guidance if the broader context of the question were presented.